SQL Exercise: Employees with s as their 3rd character in first name
SQL SORTING and FILTERING on HR Database: Exercise-19 with Solution
19. From the following table, write a SQL query to find those employees whose first name contains a character 's' in the third position. Return first name, last name and department id.
Sample table : employees
Sample Solution:
SELECT first_name,last_name, department_id
FROM employees
WHERE first_name LIKE '__s%';
Sample Output:
first_name | last_name | department_id -------------+-----------+--------------- Jose Manuel | Urman | 100 Jason | Mallin | 50 Joshua | Patel | 50 Lisa | Ozer | 80 Susan | Mavris | 40 (5 rows)
Code Explanation:
The said query in SQL which retrieves the first name, last name, and department ID columns from the 'employees' table where the value in the "first_name" column starts with two characters followed by "s".
In this case, the result will be a set of rows in the 'employees' table that match the specified pattern based on the first name.
Relational Algebra Expression:
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Previous SQL Exercise: Employees earn over 11000 or 7th digit in their phone.
Next SQL Exercise: Find employees who is working except given departments.
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SQL: Tips of the Day
Difference between natural join and inner join
One significant difference between INNER JOIN and NATURAL JOIN is the number of columns returned-
Consider:
TableA TableB +------------+----------+ +--------------------+ |Column1 | Column2 | |Column1 | Column3 | +-----------------------+ +--------------------+ | 1 | 2 | | 1 | 3 | +------------+----------+ +---------+----------+
The INNER JOIN of TableA and TableB on Column1 will return
SELECT * FROM TableA AS a INNER JOIN TableB AS b USING (Column1); SELECT * FROM TableA AS a INNER JOIN TableB AS b ON a.Column1 = b.Column1;
+------------+-----------+---------------------+ | a.Column1 | a.Column2 | b.Column1| b.Column3| +------------------------+---------------------+ | 1 | 2 | 1 | 3 | +------------+-----------+----------+----------+
The NATURAL JOIN of TableA and TableB on Column1 will return:
SELECT * FROM TableA NATURAL JOIN TableB +------------+----------+----------+ |Column1 | Column2 | Column3 | +-----------------------+----------+ | 1 | 2 | 3 | +------------+----------+----------+
Ref: https://bit.ly/3AG5CId
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